为返回采用参数和 kwarg 的函数的函数键入注释

Question

以下示例代码在运行时工作，但mypy --strict不接受：

from typing import Any, Callable, TypeVar
TypeT = TypeVar('TypeT')

def log_call(msg: str) -> Callable[..., TypeT]:
def do_call(func: Callable[..., TypeT], *args: Any, **kwargs: Any) -> TypeT:
print(msg)
return func(*args, **kwargs)
return do_call

def double(val: int) -> int:
return 2 * val

def plus(a: int, b: int) -> int:
return a + b

def plus_plus(a: int, b: int, c: int) -> int:
return a + b + c

result_1 = log_call('hi')(double, 1)
result_2 = log_call('hi')(plus, 2, 3)
result_3 = log_call('hi')(plus_plus, 2, 3, 4)
result_4 = log_call('hi')(double, val=1)
print(result_1)
print(result_2)
print(result_3)
print(result_4)

mypy输出：

test.py:26: error: Need type annotation for 'result_1'
test.py:27: error: Need type annotation for 'result_2'
test.py:28: error: Need type annotation for 'result_3'
test.py:30: error: Need type annotation for 'result_4'
test.py:32: error: Cannot determine type of 'result_1'
test.py:33: error: Cannot determine type of 'result_2'
test.py:34: error: Cannot determine type of 'result_3'
test.py:35: error: Cannot determine type of 'result_4'

现在，我想调整函数的类型注释，以便可以推断出其他符号，而不是将类型注释添加到result*变量中。这是我的尝试：

from typing import Any, Callable, TypeVar
TypeT = TypeVar('TypeT')

def log_call(msg: str) -> Callable[[Callable[..., TypeT], Any, Any], TypeT]:
def do_call(func: Callable[..., TypeT], *args: Any, **kwargs: Any) -> TypeT:
print(msg)
return func(*args, **kwargs)
return do_call

def double(val: int) -> int:
return 2 * val

def plus(a: int, b: int) -> int:
return a + b

def plus_plus(a: int, b: int, c: int) -> int:
return a + b + c

result_1 = log_call('hi')(double, 1)
result_2 = log_call('hi')(plus, 2, 3)
result_3 = log_call('hi')(plus_plus, 2, 3, 4)
result_4 = log_call('hi')(double, val=1)
print(result_1)
print(result_2)
print(result_3)
print(result_4)

但是现在参数的数量不再适合：

test.py:26: error: Too few arguments
test.py:28: error: Too many arguments
test.py:30: error: Unexpected keyword argument "val"

我想我正在寻找类似的东西

def log_call(msg: str) -> Callable[[Callable[..., TypeT], ...], TypeT]:

但这种语法是不合法的。

有没有办法解决这个问题？

Answer 1

解决此问题的最简单方法是将log_call设置为可调用对象。这会将类型注释与包装器和实际调用分开。此外，从Python 3.10开始，可以使用ParamSpec来注释输入参数(*args和**kwargs(。这是一个完整的示例，通过mypy --strict并有效：

from typing import Callable, ParamSpec
P = ParamSpec("P")

class log_call:
def __init__(self, msg: str) -> None:
self.msg = msg
def __call__(
self, func: Callable[P, int], *args: P.args, **kwargs: P.kwargs
) -> int:
print(self.msg)
return func(*args, **kwargs)

def double(val: int) -> int:
return 2 * val

def plus(a: int, b: int) -> int:
return a + b

def plus_plus(a: int, b: int, c: int) -> int:
return a + b + c

result_1 = log_call("hi")(double, 1)
result_2 = log_call("hi")(plus, 2, 3)
result_3 = log_call("hi")(plus_plus, 2, 3, 4)
result_4 = log_call("hi")(double, val=1)
print(result_1)
print(result_2)
print(result_3)
print(result_4)
print(result_4)
print(result_4)